CD and GH are respectively the bisectors of `angle ACB` and `angle EGF` such that D and H lie on sides AB and FE of `DeltaABC and Delta FEG` respectively. IF `DeltaABC~DeltaFEG` then show that
`(CD)/(GH)=(AC)/(FG)`

CD and GH are respectively the bisectors of angle ACB and angle EGF such that D and H lie on sides AB and FE of DeltaABC and Delta FEG respectively. IF DeltaABC~DeltaFEG then show that DeltaDCB~DeltaHGE

CD and GH are respectively the bisectors of angle ACB and angle EGF such that D and H lie on sides AB and FE of DeltaABC and Delta FEG respectively. IF DeltaABC~DeltaFEG then show that DeltaDCA~DeltaHGF

CD and GH are respectively the bisectors of angleACB and angleEGF such that D and H lie on sides AB and FE of DeltaABC and DeltaEFG respectively. If DeltaABC ~ DeltaFEG show that : (CD)/(GH) = (AC)/(FG)

GD and GH are respectively the bisectors of angleACB and angleEGF such that D and H lie on sides AB and FE of DeltaABC and and DeltaEFG respectively. If DeltaABC~DeltaFEG , show that: (CD)/(GH) =(AC)/(FG)

CD and GH are respectively the bisectors of angleACB and angleEGF such that D and H lie on sides AB and FE of DeltaABC and DeltaEFG respectively. If DeltaABC ~ DeltaFEG show that : DeltaDCB~ DeltaHGE

GD and GH are respectively the bisectors of angleACB and angleEGF such that D and H lie on sides AB and FE of DeltaABC and and DeltaEFG respectively. If DeltaABC~DeltaFEG , show that: DeltaDCB ~DeltaHGE

GD and GH are respectively the bisectors of angleACB and angleEGF such that D and H lie on sides AB and FE of DeltaABC and and DeltaEFG respectively. If DeltaABC~DeltaFEG , show that: DeltaDCA~DeltHGF

CD and GH are respectively the bisectors of /_A C B and /_E G F such that D and H lie on sides AB and FE of DeltaA B C and DeltaE F G respectively. If DeltaA B C ~DeltaF E G , show that: (i) (C D)/(G H)=(A G)/(F G) (ii) ∆ DCB ~ ∆ HGE (iii) ∆ DCA ~ ∆ HGF

CD and GH are respectively the bisectors of /_ACB" and "/_EGF such that D and H lie on sides AB and FE of DeltaABC" and "DeltaEFG respectively. If DeltaABC ~ DeltaFEG , show that : (CD)/(GH)=(AC)/(FG)

CD and GH are respectively the bisectors of angleACB and angleEGF such that D and H lie on sides AB and FE of triangleABC and triangleEFG respectively. If triangleABC ~ triangleFEG , show that :- (CD)/(GH)=(AC)/(FG) .